Question

解决我没有使用＆＃34; name＆＃34;在输入字段中。当我为输入添加名称时，我立即工作。

我有一个中等大小的菜单，里面有几个子类别和很多复选框。我使用php，mysql和jquery / javascript开发自定义CMS。

菜单采用带按钮的形式＆＃34; Save＆＃34;需要从选中的复选框中获取所有值。

在按钮上，我使用onclick函数指向我的班级，在该班级中我使用这些id并将它们存储在表格中。

<input type="submit" value="Save" class="save-butt" onclick="sjx('domain', $( ':checkbox' ).serializeArray()); return false;">

它不起作用;我没有收到任何错误，但似乎没有发送该值。如果它有帮助，这是我的菜单代码。

          <div class="personal-right-box">
        <ul class="doe-list">

        <?php 
        $mainCat = Db::query('SELECT * FROM categories WHERE parent_id = 0 ORDER BY orderby DESC');
        foreach ($mainCat as $main) { ?>

          <li>
            <span><i class="fa fa-caret-down toggle-down"></i><?php echo $main['title_hr']; ?></span>
            <ul class="doe-list-sub">
              <?php 
              $subCat = Db::query('SELECT * FROM categories WHERE parent_id = '.$main['id'].' ORDER BY title_hr ASC');
              foreach ($subCat as $sub) { ?>
              <li>                    
                <input type="checkbox" name="" id="aaeroplanes-b1-b2" value="<?php echo $sub['id']; ?>">
                <label for="aaeroplanes-b1-b2"><?php echo $sub['title_hr']; ?></label>
                <i class="fa fa-caret-down toggle-down"></i>
                <ul class="doe-list-sub">
                  <?php 
                  $subsubCat = Db::query('SELECT * FROM categories WHERE parent_id = '.$sub['id'].' ORDER BY title_hr ASC');
                  foreach ($subsubCat as $subsub) { ?>
                  <li>
                    <input type="checkbox" name="" id="piston-engine-aeroplanes-b1-2" value="<?php echo $subsub['id']; ?>">
                    <label for="piston-engine-aeroplanes-b1-2"><?php echo $subsub['title_hr']; ?></label>
                    <i class="fa fa-caret-down toggle-down"></i>
                    <ul class="doe-list-sub">
                      <?php 
                      $subsubsubCat = Db::query('SELECT * FROM categories WHERE parent_id = '.$subsub['id'].' ORDER BY title_hr ASC');
                      foreach ($subsubsubCat as $subsubsub) { ?>
                      <li>
                        <input type="checkbox" name="" id="cessna" value="<?php echo $subsubsub['id']; ?>">
                        <label for="cessna"><?php echo $subsubsub['title_hr']; ?></label>
                        <i class="fa fa-caret-down toggle-down"></i>
                        <ul class="doe-list-sub group">
                          <?php 
                          $items = Db::query('SELECT * FROM categories WHERE parent_id = '.$subsubsub['id'].' ORDER BY title_hr ASC');
                          foreach ($items as $item) { ?>
                          <li class="col-1-3">
                            <input type="checkbox" name="" id="" value="<?php echo $item['id']; ?>">
                            <label for=""><?php echo $item['title_hr']; ?></label>
                          </li>
                          <?php } ?>
                        </ul>
                      </li>
                      <?php } ?>
                    </ul>
                  </li>
                  <?php } ?>
                </ul>
              </li>
              <?php } ?>
            </ul>
          </li>

        <?php } ?>
        </ul><!-- end of .doe-list -->
        <span class="save-button-form">
          <input type="submit" value="Save" class="save-butt" onclick="sjx('domain', $( ':checkbox' ).serializeArray()); return false;">
        </span>
      </div>

感谢您的帮助。

Answer 1

序列化首先使用name，如果找不到，则会转到id。在您的情况下，您有一个空白name，结果也是如此。
您可以在name中复制相同的内容或将其删除。

使用.serializeArray

1 个答案: